Principal Unit Normals from Tangent Derivatives

From a Changing Tangent to a Normal Direction

A unit tangent vector records the direction in which a regular curve is moving. As the parameter changes, that direction may change too. The derivative of the unit tangent points in the direction of this change. When this derivative is nonzero, its direction defines the curve's principal unit normal.

The construction has an important limitation: if the derivative of the unit tangent is zero at a point, it cannot be normalized there. The principal unit normal defined by this method is then undefined at that point. A curve may still have a well-defined tangent there.

Throughout, let $\mathbf r(t)$ be a twice-differentiable regular curve on an interval, so that $\mathbf r'(t)\ne\mathbf 0$. Its unit tangent is $\mathbf T(t)=\mathbf r'(t)/\|\mathbf r'(t)\|$. The earlier material on unit tangent vectors establishes this formula for non-unit-speed parametrizations.

Definition: At a parameter value $t$ where $\mathbf T'(t)$ exists and is nonzero, the principal unit normal is $$ \mathbf N(t)=\frac{\mathbf T'(t)}{\|\mathbf T'(t)\|}. $$ If $\mathbf T'(t)=\mathbf 0$, this definition does not give a principal unit normal at $t$.

The order of operations matters: first find the unit tangent, then differentiate it, and only then divide by the length of that derivative. In general, $\mathbf T'(t)$ is not itself a unit vector.

Why the Normal Is Perpendicular to the Tangent

The unit tangent has constant length $1$. Differentiating the squared length shows that its derivative is perpendicular to it. This explains why the direction obtained by normalizing $\mathbf T'$ is a normal direction.

Derivation: Since $\mathbf T(t)\cdot\mathbf T(t)=1$, differentiating both sides gives $$ \frac{d}{dt}\big(\mathbf T(t)\cdot\mathbf T(t)\big) =2\mathbf T(t)\cdot\mathbf T'(t)=0. $$ Thus $\mathbf T(t)\cdot\mathbf T'(t)=0$. Wherever $\mathbf T'(t)\ne\mathbf 0$, division by its length preserves this perpendicularity, so $\mathbf T(t)\cdot\mathbf N(t)=0$. Also, the definition directly gives $\|\mathbf N(t)\|=1$.

In a space curve, there are many directions perpendicular to a given tangent. This construction selects one specific direction: the direction in which the unit tangent is changing. If the tangent is not changing at a particular parameter value, that selected direction is unavailable there.

A Reliable Calculation Process

Find the unit tangent.
Compute $\mathbf r'(t)$, its speed $\|\mathbf r'(t)\|$, and $\mathbf T(t)=\mathbf r'(t)/\|\mathbf r'(t)\|$.

Differentiate the unit tangent.
Use the derivative rules on every component of $\mathbf T(t)$.

Check where the derivative vanishes.
If $\mathbf T'(t)=\mathbf 0$, the principal unit normal is undefined at that parameter value.

Normalize where possible.
For $\mathbf T'(t)\ne\mathbf 0$, compute its length and divide each component by that length.

Verify the result.
Check that the proposed normal has length $1$ and has dot product $0$ with $\mathbf T(t)$.

Worked Examples

Worked Example: A Parabola in the Plane

Find the principal unit normal of $\mathbf r(t)=\langle t,t^2\rangle$, and evaluate it at $t=1$.

Differentiate the curve and calculate its speed:

\mathbf r'(t)=\langle 1,2t\rangle, \qquad \|\mathbf r'(t)\|=\sqrt{1+4t^2}.

Therefore the unit tangent is

\mathbf T(t)=\frac{\langle 1,2t\rangle}{\sqrt{1+4t^2}} =\left\langle (1+4t^2)^{-1/2},\,2t(1+4t^2)^{-1/2}\right\rangle.

Differentiate the first component. Differentiating the second by the product rule gives

\begin{aligned} T_1'(t)&=-4t(1+4t^2)^{-3/2},\\ T_2'(t)&=2(1+4t^2)^{-1/2}-8t^2(1+4t^2)^{-3/2}\\ &=\frac{2(1+4t^2)-8t^2}{(1+4t^2)^{3/2}} =\frac{2}{(1+4t^2)^{3/2}}. \end{aligned}

Hence

\mathbf T'(t)=\frac{\langle-4t,2\rangle}{(1+4t^2)^{3/2}}, \qquad \|\mathbf T'(t)\| =\frac{\sqrt{16t^2+4}}{(1+4t^2)^{3/2}} =\frac{2}{1+4t^2}.

This length is positive for every real $t$. Normalize to obtain

\mathbf N(t) =\frac{\langle-4t,2\rangle}{2\sqrt{1+4t^2}} =\frac{\langle-2t,1\rangle}{\sqrt{1+4t^2}}.

At $t=1$, $\mathbf N(1)=\langle-2,1\rangle/\sqrt5$. Its length is $\sqrt{4+1}/\sqrt5=1$. Also, $\mathbf T(1)=\langle1,2\rangle/\sqrt5$, and their dot product is $(-2+2)/5=0$. Both checks hold.

Worked Example: A Circle

For the circle $\mathbf r(t)=\langle4\cos t,4\sin t,0\rangle$, find the principal unit normal and evaluate it at $t=\pi/3$.

Differentiate and find the speed:

\mathbf r'(t)=\langle-4\sin t,4\cos t,0\rangle, \qquad \|\mathbf r'(t)\|=4.

Thus $\mathbf T(t)=\langle-\sin t,\cos t,0\rangle$, and differentiating gives

\mathbf T'(t)=\langle-\cos t,-\sin t,0\rangle, \qquad \|\mathbf T'(t)\|=\sqrt{\cos^2t+\sin^2t}=1.

Since the derivative is nonzero for every $t$, the principal unit normal is

\mathbf N(t)=\langle-\cos t,-\sin t,0\rangle.

At $t=\pi/3$, this is $\langle-1/2,-\sqrt3/2,0\rangle$, whose length is $\sqrt{1/4+3/4}=1$. The tangent there is $\langle-\sqrt3/2,1/2,0\rangle$. Their dot product is $\sqrt3/4-\sqrt3/4=0$, as required.

Worked Example: A Point Where the Normal Is Undefined

Consider $\mathbf r(t)=\langle t,t^3,0\rangle$. Find the principal unit normal wherever it exists, and determine what happens at $t=0$.

The velocity and speed are

\mathbf r'(t)=\langle1,3t^2,0\rangle, \qquad \|\mathbf r'(t)\|=\sqrt{1+9t^4}.

The speed is positive for every $t$, so the curve is regular. Its unit tangent is

\mathbf T(t)=\frac{\langle1,3t^2,0\rangle}{\sqrt{1+9t^4}}.

Differentiating the first component gives $-18t^3/(1+9t^4)^{3/2}$. For the second component, the product rule gives

\begin{aligned} \frac{d}{dt}\left(3t^2(1+9t^4)^{-1/2}\right) &=6t(1+9t^4)^{-1/2}-54t^5(1+9t^4)^{-3/2}\\ &=\frac{6t(1+9t^4)-54t^5}{(1+9t^4)^{3/2}} =\frac{6t}{(1+9t^4)^{3/2}}. \end{aligned}

Therefore,

\mathbf T'(t)=\frac{6t\langle-3t^2,1,0\rangle}{(1+9t^4)^{3/2}}, \qquad \|\mathbf T'(t)\| =\frac{6|t|\sqrt{9t^4+1}}{(1+9t^4)^{3/2}} =\frac{6|t|}{1+9t^4}.

For $t\ne0$, normalization yields

\mathbf N(t) =\frac{t}{|t|}\frac{\langle-3t^2,1,0\rangle}{\sqrt{1+9t^4}} =\frac{\operatorname{sgn}(t)\langle-3t^2,1,0\rangle}{\sqrt{1+9t^4}}.

At $t=0$, $\mathbf T'(0)=\mathbf0$, so $\mathbf N(0)$ is undefined by this construction. The unit tangent itself is defined there: $\mathbf T(0)=\langle1,0,0\rangle$. As $t$ approaches zero from the right, $\mathbf N(t)$ approaches $\langle0,1,0\rangle$; from the left, it approaches $\langle0,-1,0\rangle$. Thus there is not even a single two-sided limiting value to use as a continuous extension.

For a check at $t=2$, the formula gives $\mathbf N(2)=\langle-12,1,0\rangle/\sqrt{145}$, which has length $1$. The tangent there is $\langle1,12,0\rangle/\sqrt{145}$, and the dot product is $(-12+12)/145=0$.

A Common Normalization Error

In the preceding example, it is easy to mishandle the square root when computing $\|\mathbf T'(t)\|$. The correct length is $6|t|/(1+9t^4)$, not $6|t|/\sqrt{1+9t^4}$. The factor $\sqrt{1+9t^4}$ in the numerator of the length cancels only one of the three powers in $(1+9t^4)^{3/2}$.

If one uses the incorrect length $6|t|/\sqrt{1+9t^4}$, division of the actual $\mathbf T'(t)$ by that incorrect quantity gives $\operatorname{sgn}(t)\langle-3t^2,1,0\rangle/(1+9t^4)$. This is not a unit vector: its length is $1/\sqrt{1+9t^4}$. For example, at $t=2$, the result is $\langle-12,1,0\rangle/145$, with length $1/\sqrt{145}$, not $1$.

A dependable check is to calculate the length of the final proposed normal directly. A candidate that is perpendicular to the tangent but does not have length $1$ is a normal direction, not a unit normal.

What Undefined Means Here

When $\mathbf T'(t)=\mathbf0$, the definition requires division by $\|\mathbf T'(t)\|=0$, which is not allowed. This says that the principal unit normal from tangent differentiation is undefined at that parameter value. It does not say that the curve is not regular: in the example above, $\mathbf r'(0)\ne\mathbf0$, so the tangent exists even though the principal unit normal does not.

Likewise, a formula for $\mathbf N(t)$ valid away from a point must not automatically be used at that point. Check the definition there by evaluating $\mathbf T'(t)$. If it vanishes, a separately defined limiting direction, even if one exists from one side, is not the value supplied by normalization.

Check Your Understanding

Use the definition and the checks in this tutorial to answer the following.

For a regular curve, what two operations turn the unit tangent into the principal unit normal?
Why is $\mathbf T'(t)$ perpendicular to $\mathbf T(t)$ wherever the derivative exists?
What condition on $\mathbf T'(t)$ is required to define $\mathbf N(t)$ by normalization?
For $\mathbf T(t)=\langle\cos t,\sin t,0\rangle$, find $\mathbf N(t)$ and check its length.
Can a regular curve have a defined unit tangent but an undefined principal unit normal at a point? Explain using the defining condition.